A wheel is rolling with the velocity of its center equal to V. At some time, a small particle is placed at the top of the wheel such that there is no relative motion between the rim of the wheel and the particle. Find the velocity V,such that the particle will immediately fly away the moment it is placed. The radius of wheel is R.

Question

anonymous · Answer

|dw:1372336491449:dw|

The answer is sqrt(gR)

anonymous · Answer

I know that the particle would fly off if the normal rxn with the rim is zero. The weight and the normal rxn should provide it the centripetal force. The velocity of rim is 2V and since the relative motion is not there, the particle was released with velocity 2V. But i do not know about what point the particle is undergoing circular motion.

anonymous · Answer

i'm nt sure... but.... i thnk de particle experiences circular motion at de point before it is released.... der 
$$mg \cos \theta = mv^2/ R$$

theeric · Answer

I apologize now if I am unclear - it's a little past my bed time (I guess that happens to me). When will the particle immediately leave the wheel? When there is no friction. None, or else it would not be immediate. But there was no mention of "frictionless," so there must be no friction for a different reason: no normal force. You mentioned the normal force, I think! No normal force? Must be because the particle does not have a NET force on the wheel! Why? Centrifugal force counteracts gravitational!|dw:1372490603077:dw|So we know that$$\large F_{centrifugal}=F_{gravitational}$$ And the centrifugal force is just opposite in magnitude to centripetal.$$F_{centrifugal}=-F_{centripetal}=-\frac{mv^2}{r}$$ http://hyperphysics.phy-astr.gsu.edu/hbase/corf.html#cent $$F_{gravity}$$you might know, is$$mg$$$$\text{where }g \text{ is gravitational acceleration!}$$ By substitution into $$\large F_{centrifugal}=F_{gravitational}$$we get$$-\frac{mv^2}{r}=mg$$ We can cancel like variables and, overall, solve for v.$$-\frac{\cancel{m}v^2}{r}=\cancel{m}g$$$$-\frac{v^2}{r}=g$$$$\Downarrow \ v^2=-rg\\Downarrow\\sqrt{-rg}$$

theeric · Answer

Negative under radical? Nope!  negative, so the negatives cancel and the radicand is positive. I think that's what it's called.

I hope this helps! It's finding equations by physics and solving with math, overall. It gets to be a common practice!

anonymous · Answer

not quite satisfied. But thanks for trying!